Question

The goal is to integrate from 0 to 1/16 for `int_0^1(arcsin(8x))/(sqrt(1-64x^2)) dx` The problem I'm encountering is the when I make my second substitution (kind of like recursion) to change the integrand on this definite intergral, I end up having to evaluate the arcsin of zero (which is zero, no problem) and the arcsin of 128 (which a normal graphic calculator like a TI-84 freaks out an says it's "Out of Domain" whatever that means). Am I doing something incorrect?

My first substitution is u = 8x , thus du/8 = dx
My second subsitution is w = arcsin(u), thus dw = `(1)/(sqrt(1-u^2))`

Answer 1

PS: I don't think the syntax for the MathJax or ASCIIMathML is showing correctly, I saw somebody else get it to work previously. The correct answer I believe is something like 0.017... but what I'm needing is an exact answer which more than likely has π (Pi) in it somewhere.

Hopefully this syntax shows the integral correctly now:
\[\int\limits_{0}^{1/16} (\sin^{-1} (8x)) / (\sqrt(1-64x^2))\]

and the second substitution so it's easier to see
dw = \[1/\sqrt(1-u^2)\]

Answer 2

that's funny because my TI-84 shows 0.017 as the answer also...

Answer 3

Well I finally got the solution using this substitution instead. If I only could remember what arcsin of of 0.5 was... then I'd have the exact solution for the definite integral.
\[u = \sin^{-1} (8x)\]
\[du = 8/(1-(8x)^2) --> (1/8)du = 1/(1-64x^2)\]

Answer 4

arcsin(.5) = pi/6